Question

Tìm x biết: (x-3)^2-16=0

x^2 -2x=24

(2x+1)+(x+3)^2(x+7)(x-7)=0

Answer 1

x^2 -2x = 24

=> x^2 - 2x - 24=0

=>x^2 -8x+6x - 24 = 0

=> ( x^2- 8x)+( 6x-24) = 0

=> x(x-8) + 6(x-8) = 0

=> (x+6)(x-8)=0

=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)

Answer 2

\(=\frac{\left(2.5\right)^4.3^4-2^4\left(3.5\right)^2}{2^8.5^2.3^3}=\frac{2^4.3^2.5^2\left(5^2.3^2-1\right)}{2^8.5^2.3^3}=\frac{255-1}{16.3}=\frac{14}{3}\)

Answer 3

xin lỗi mình làm lộn bài

x=6

Answer 4

        (x-3)^2 - 16 = 0

=> (x-3)^2 = 16

=> \(\orbr{\begin{cases}x-3=4\\x-3=-4\end{cases}}\)

\(=>\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)

Answer 5

_ \(\left(x-3\right)^2-16=0\Rightarrow\left(x-3\right)^2=16\Rightarrow x-3=4\Rightarrow x=4+3=7\)
_   \(x^2-2x=24\Rightarrow x^2-2x-24=0\)
    \(\Rightarrow x^2-2x+1-25=0\Rightarrow\left(x^2-2x+1\right)-25=0\)
    \(\Rightarrow\left(x-1\right)^2-5^2=0\Rightarrow\left(x-1+5\right)\left(x-1-5\right)=0\)
    \(\Rightarrow\left(x+4\right)\left(x-6\right)=0\)
    \(\Rightarrow\orbr{\begin{cases}x+4=0\Rightarrow x=-4\\x-6=0\Rightarrow x=6\end{cases}}\)
Vậy x= (- 4) và 6
_ CÂu này mik chưa nghĩ ra^^!